Combinations of samples problem

[jonroberts74]

Homework Statement

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.
a. How many different samples can be chosen?
b. How many samples will contain at least one defective board?
c. What is the probability that a randomly chosen sample of five contains at least one defective board?

The Attempt at a Solution

A) $\binom{40}{5} = \frac{40!}{5!(35!)} = 658\;008$

B) $\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{25}{658\;008}$

 

[jonroberts74]

B) $\binom{3}{1}\binom{37}{4}+\binom{3}{2}\binom{37}{3}+\binom{3}{3} \binom{37}{2} = 222\,111$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{222\,111}{658\,008} \approx 33.76\%$

 

[vela]

Your second attempt looks good.

Another way to calculate (B) is to figure out how many ways to select 5 good boards and then subtract that number from your answer to (A).